Tree Selection
Jun 16, 2008
Thank You visakh16 --- The following section of my querie works but doesn't give the exact information that I need. This is what is giving me:
NetworkId Full Name Sec Class Desc Sec Type Desc SecValue
tte Test Scenario Accounting Budget Center 142- ?
SELECT a.Network_ID, b.Last_Name + ', ' + b.Preferred_FirstName AS full_name, c.Security_Class_Description,
d.Security_Type_Description, a.Security_Value
FROM Company_Hierarchy_Security a
JOIN V_Entity_Employee_Active b on a.Network_ID= b.Network_ID
JOIN Company_Security_Class c on a.Security_Class_Code=c.Security_Class_Code
JOIN Company_Security_Type d on a.Security_Type = d.Security_Type
------------------
inner join (select e.Budget_Center_ID + ' - ' + e.Budget_Center_Description As Budget_Center_Description,
f.Company_Name, g.Enterprise_Description, h.Business_Segment_Description,
i.Team_Description
from Company_Hierarchy_Security a.
Inner JOIN Budget_Center e on a.Security_Value = e.Budget_Center_ID
Inner JOIN Company f on a.Security_Value = f.Company_ID
Inner JOIN Enterprise g ON a.Security_Value = Cast(g.Enterprise_Number As Varchar(5))
Inner JOIN Business_Segment h on a.Security_Value = h.Business_Segment_ID
Inner JOIN Team i on a.Security_Value = i.Team_ID
_____________________________________________________________________
What I need is the Description that are located in 4 other different tables that matches the Security Value from my first querie.
The result should look like this...
Network Id full Name Sec Class Desc Sec Type Desc Security Value
tst , Test Example , Accounting ,Budget Center , 142-Accountig dept
Thank you very much,
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Jun 13, 2008
SELECT a.Network_ID, b.Last_Name + ', ' + b.Preferred_FirstName AS full_name, c.Security_Class_Description,
d.Security_Type_Description, a.Security_Value
FROM Company_Hierarchy_Security a
JOIN V_Entity_Employee_Active b on a.Network_ID= b.Network_ID
JOIN Company_Security_Class c on a.Security_Class_Code=c.Security_Class_Code
JOIN Company_Security_Type d on a.Security_Type = d.Security_Type
inner join (select e.Budget_Center_ID + ' - ' + e.Budget_Center_Description As Budget_Center_Description,
f.Company_Name, g.Enterprise_Description, h.Business_Segment_Description,
i.Team_Description
from Company_Hierarchy_Security a.
Inner JOIN Budget_Center e on a.Security_Value = e.Budget_Center_ID
Inner JOIN Company f on a.Security_Value = f.Company_ID
Inner JOIN Enterprise g ON a.Security_Value = Cast(g.Enterprise_Number As Varchar(5))
Inner JOIN Business_Segment h on a.Security_Value = h.Business_Segment_ID
Inner JOIN Team i on a.Security_Value = i.Team_ID
Ok. I have the Security Value located in Company_Hierarchy_Security table. All those values are divided into 5 other tables that I need to join together. I found all of them separate but I have not been able to figue it out how to put it together with the rest of the querie.
Thanks for the help!!!!!
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Apr 6, 2015
I got assignment, how to make it appear in the right order .
/* DROP TABLE EMP
SELECT * INTO Emp FROM (
SELECT 'A' EmpID, NULL ManID, 'Name' EmpName UNION ALL
SELECT 'MAC' EmpID, 'A' ManID, 'Name__' EmpName UNION ALL
SELECT '1ABA' EmpID, 'MAC' ManID, 'Name____' EmpName UNION ALL
SELECT 'ABB' EmpID, '1ABA' ManID, 'Name______' EmpName UNION ALL
SELECT 'XB' EmpID, 'A' ManID, 'Name__' EmpName UNION ALL
SELECT 'BAC' EmpID, 'XB' ManID, 'Name____' EmpName ) b
*/
[code]....
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May 20, 2004
Dear all,
I would like to know how to create a single level hierachy structure in SQL.
Example, I have a single parent record in table A that may later on spawn one or child record in table B that relates back to the parent. It only needs a single level.
that means, one to many.
thanks and rgds.
Loke HC
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Jan 2, 2008
I have two tables.
table1 has 3 fields a , b and c. field "a" is a primary key.
table2 has 2 fields x and y. Fields x and y are nothing but the value of "a". Also, y is the child of x.
Therefore, x and y can never have same value. It means value of "a" either be child or parent.
But there is possibility that parent has no child.
Now, i wanted to write Select/Insert query for parent, b , c and child.
This tree is no a binary tree but N-Ary Tree.
Thanks,
sha
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May 18, 2007
Hi,
I'm having a major brain-failure moment here.
Using T-SQL I want to be able to get all of the leaf nodes (e.g. nodes at the furthest end of a tree from the root) in a hierarchical relationship where the table structure is such NodeID|ParentID|NodeName.
Basically if I had the following tree structure:
Root
Child1
Grandchild1
Child2
Grandchild2
I want to get all of the Grandchild nodes. Number of levels will vary and I haven't got any kind of HasChilds column.
I know this is possible because I remember having done it on a course years ago but I can't for the life of me figure it out on this sunny Friday afternoon. I know it's going to involve either recursion, a while loop or cursors but my mind is currently jelly. Can anyone help?
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Sep 10, 2007
hi. I am working on Multiline Marketing Project.I have to calculate all the childens and display the whole tree..I have save records in tree format...But not able to calculate the all childrens.
SELECT count(*) FROM dfTree WHERE id in (SELECT id FROM dfTree WHERE lineage like '16%')
This query works properly.Problem is that it is not working in the project because in the like I have to pass a variable.
SELECT count(*) FROM dfTree WHERE id in (SELECT id FROM dfTree WHERE lineage like '@idl%')
Please Suggest me...
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Jun 2, 2008
HI,
I am working on a Family tree portal which need tree functionality to display family members in tree structure. on click on any node the adding option should be displayed
for this i need a table and procedure to complete family tree
Thanks
@mbi
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Jan 3, 2006
Consider the following SQL query:
SELECT ENAME,SAL
FROM EMP,ASG,PAY,PROJ
WHERE EMP.ENO=ASG.ENO
AND PAY.TITLE=EMP.TITLE
AND ASG.PNO=PROJ.PNO
AND ASG.DUR=48 AND BUDGET>200000
Give the possible operator trees:right-deep,left-deep and bush
Tank you very much!
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Jun 11, 2006
Hi,
i'm writing a app in c# and have to store Trees in a Database.
I'm working with Datasets for the exchange between the DB and the App.
The trees have the same options like the windows folders. If u delete a node, all subnodes should be deleted too.
But something a Foreign Key from ParentID references (Id) with the delete-Rule on cascade seems not to be possible, because of multiple cascade Paths or cycles. Do i have to add some xtra constarins:
Not Possible:
create Table tree (
Id varchar Not null,
ParentId varchar Not null,
Constraint pk1 Primary Key (Id),
Constraint fk1 Foreign Key (ParentId) references tree(Id)
On Update Cascade
On delete CAscade
)
Do i have to write triggers, which delete The subnodes too and set the Update-/deleterulr on NO Action
Greetz
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Oct 24, 2006
Does anyone know any good links for SQL tree structures and example queries and stuff... I cant really find anything part from the standard example of emplyee, boss, salary which explains how to create the tree table...(dun dis bit) I did notice a book but I live in a little village so cant go get it till wekend?
I'm desperate, reli need to work out how too do this.....
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Oct 14, 2005
Any recomendations on how to store organization trees on a database withlots of paths and branches? Any white papers out there that explain this?Thanks
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Sep 12, 2006
Hello,I have a "tree" table:Id - primary keyParrentId - (foreign key) related to IdTitle.....when I delete some record I want to delete it with all childs (cascadedeleting). I can't set cascade deleting on the same table :(. Is thereany easy way in the MSSQL 2005 to do this ? There is one idea - usingcursors + recursive functions but I think this solution is not easyand elegant.Thakns for any help and sugestions.Regards.Andy
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Mar 5, 2008
Hi:
Is there a resource or an object in Reporting Services that I could make a tree view in a report file .rdl and see the result on it?
If not, is there a software of MS that a could make a component for it, and connect this component in the report file .rdl ?
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Apr 16, 2007
I'm having this problem.....
I wanted to use the Decision Tree to show a result..... after i configure the Mining Structures..... and set all the input.... my decision tree shows only until level 2..... i have 3 input and one PredictOnly column.....where is the other input?
Say.... i have House Owner, Marital Status, Num Cars Owned and Number Of Children(PredictOnly)
my Tree only shows All ---- > Marital Status when i input all 3 together...... the other 2 doesn't seems to show.
wat should i do?? my database in SQL Server and the other keys are all correct and deploying finely.....why is this happening.....?
i'm a newbie in this software.......so any pro here can plz help me if there's actually something that i might have missed out along the way.......
Thank you again.........
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Feb 28, 2008
Alright what im trying to do is build a classic tree, i have id's like this set up in my table referencing itself in sql 2000
cmponent_prt_no parent_part_no
--------------- --------------
112837A2A L115100-1
114379A1A L115100-1
115623A1A L115100-1
203604A L115100-1
203790A L115100-1
203791A L115100-1
115623A 115623A1A
M010137 115623A1A
115623A 115623A1A
20766CR 115623A1A
DRAWINGS 20766CR
M010137 20766CR
i have tried 3 different solutions all coming to a crash at some point, if ANYONE can help me out i would appreciate it, i know the logic will be recursive, i just dont know how to implement
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Aug 21, 2007
Hi..I have a table register..in this fields are username,parent id,downline ; I have to determine all the child of a particular parent.
suppose table is like this. username parentid downline
B A left
C A right
D B left
E B right....
I have to also determine the level in the tree....please help...
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Dec 6, 2000
Hello all!
I have a table defined as:
create table OrgTree (parent int, child int)
OrgTree has the following data:
1,2
1,3
2,4
2,5
3,6
4,7
4,8
8,9
Is there a SQL stmt that will return, given a particlar parent, all of the children (and children's children, etc.) of that parent?
So, parent value 1 returns 2,3,4,5,6,7,8,9
parent value 2 returns 4,5,7,8,9
parent value 3 returns 6
etc.
Thanks in advance for your help!!
Palmer F
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Oct 30, 2001
I just started having this problem. When I log into SQL2000 enterprise manager and click on my database, the main directory tree shows up. When I click on the Database tree to display all of the different databases on the SQL server, the text 'No Items' shows up. I can not get to any databases, but I am connected to the SQL server. Does anyone have a solution for this problem?????
I have tried re-inatalling and all of the service packs.
Thanks !!
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Jul 26, 2004
Okie, this one has me a little stumped so bear with me an I will explain as best I can....
I have a family tree database...
The two key tables (for this problem) are Person and Child. The person table is your generic person details table and contains, PersonId, PersonFirstName, PersonLastName, PersonDOB etc. The Child table is a linking table that links a PersonId to a ChildPersonId (so both Id's are valid Id's in the person table).
Now what I want to try and achieve is a breakdown of the different branchs of the family tree. A branch is an independant, unattached section of the tree (if that makes sense). It's a grouping of people we know are related but we can't determine how they are related to another group of people in the tree.
If you look at this http://gw.geneanet.org/index.php3?b=bengos&lang=en;m=N;v=van+halewyn you will get an idea of what I mean.
I'm not sure if this is something that can be don't with a query at all the be honest... I suspect that I will have to wrap some other code around it, but I'm really not sure on what approach I should be using. Any help people could offer would be great.
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Jul 8, 2007
Hi,
table(id,parent,level, name)
parent ...fk referenced table(id))
level ... depth of tree (root has 0)
How SQL for hiearchical list:
name 1
name 1.1
name 1.2
name 2.
name 2.1.
name 2.1.1
name 2.1.2
etc.
???
thanks for your help
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Sep 17, 2007
One of the users on another web site posted a question about how to associate users in a tree-like organization. That web site isn't well suited to posting code or ongoing discussions about code, so I'm going to post the example here. Feel free to discuss as you see fit.DROP TABLE LI_UserLinks
GO
DROP TABLE LI_Users
GO
DECLARE @d1DATETIME
, @d2DATETIME
, @d3DATETIME
, @d4DATETIME
, @d5DATETIME
, @d6DATETIME
, @d7DATETIME
, @d8DATETIME
SELECT @d1 = GetDate()
CREATE TABLE LI_Users (
uidINT
PRIMARY KEY (uid)
)
SELECT @d2 = GetDate()
CREATE TABLE LI_UserLinks (
uid_fromINT
CONSTRAINT XFK01LI_UserLinks FOREIGN KEY (uid_from)
REFERENCES LI_Users (uid)
, uid_toINT
CONSTRAINT SFK01LI_UserLinks FOREIGN KEY (uid_to)
REFERENCES LI_Users (uid)
CONSTRAINT XPKLI_UserLinks PRIMARY KEY (uid_from, uid_to)
)
ALTER TABLE LI_Userlinks
ADD CONSTRAINT XCK01LI_UserLinks CHECK (uid_from != uid_to)
SELECT @d3 = GetDate()
INSERT INTO LI_Users (
uid) SELECT n0 + 10 * n1 + 100 * n2 + 1000 * n3 + 10000 * n4 + 100000 * n5
FROM (SELECT 0 AS n0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) AS z0
CROSS JOIN (SELECT 0 AS n1 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) AS z1
CROSS JOIN (SELECT 0 AS n2 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) AS z2
CROSS JOIN (SELECT 0 AS n3 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) AS z3
CROSS JOIN (SELECT 0 AS n4 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) AS z4
CROSS JOIN (SELECT 0 AS n5 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) AS z5
SELECT @d4 = GetDate()
INSERT INTO LI_UserLinks (
uid_from, uid_to) SELECT
uid, 100 * uid + n0 + 10 * n1
FROM LI_Users
CROSS JOIN (SELECT 0 AS n0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) AS z0
CROSS JOIN (SELECT 0 AS n1 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) AS z1
WHERE LI_Users.uid BETWEEN 0 AND 99
AND uid != 100 * uid + n0 + 10 * n1
SELECT @d5 = GetDate()
INSERT INTO LI_UserLinks (
uid_from, uid_to) SELECT
uid, 100 * uid + n0 + 10 * n1
FROM LI_Users
CROSS JOIN (SELECT 0 AS n0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) AS z0
CROSS JOIN (SELECT 0 AS n1 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9) AS z1
WHERE LI_Users.uid BETWEEN 100 AND 9999
SELECT @d6 = GetDate()
SELECT u.uid, r1.uid_to, r2.uid_to
FROM LI_Users AS u
INNER JOIN LI_UserLinks AS r1
ON (r1.uid_from = u.uid)
INNER JOIN LI_UserLinks AS r2
ON (r2.uid_from = r1.uid_to)
WHERE 1 = u.uid
SELECT @d7 = GetDate()
SELECT Count(DISTINCT u.uid), Count(DISTINCT r1.uid_to), Count(distinct r2.uid_to)
FROM LI_Users AS u
INNER JOIN LI_UserLinks AS r1
ON (r1.uid_from = u.uid)
INNER JOIN LI_UserLinks AS r2
ON (r2.uid_from = r1.uid_to)
SELECT @d8 = GetDate()
SELECT
DateDiff(ms, @d1, @d2)
, DateDiff(ms, @d2, @d3)
, DateDiff(ms, @d3, @d4)
, DateDiff(ms, @d4, @d5)
, DateDiff(ms, @d5, @d6)
, DateDiff(ms, @d6, @d7)
, DateDiff(ms, @d7, @d8)
, DateDiff(ms, @d1, @d8)-PatP
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Sep 27, 2013
Here is the table - Company with fields:
CompanyID, ParentCompanyID (both integers)
Given a CompanyID - I want to get all the children for the Company.
I did similar procedures but somehow, could not get this to work.
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Feb 9, 2015
I have following 4 rows in a table with field name as ID
ID
--
AA
BB
CC
DD
I require output as
ID NewColumn
-- ----------
AA AA
BB AA~BB
CC AA~BB~CC
DD AA~BB~CC~DD
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Apr 16, 2007
I have a tree structured table like the ff.
which looks like the one below in tree view.
I'd like to get the rows under a specified elementnumber which may have unlimited sub-nodes.
i.e. If i want only 'Level1' i would get results from row 2-5.
any help is greatly appreciated.
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Jul 20, 2005
hello i have implemented joe celko's model to store heirarchy and itworks well and nicei have a question about SQLthis is the actual tablemember side left right------------------------------------------nancy L 1 36andrew L 4 21steven R 5 12ina L 6 7david R 10 11margaret L 13 20ann R 14 15laura L 18 19janet R 24 35michael L 25 30dan R 26 27ron L 28 29robert R 33 34the Side column is to tell its left, or right. this is a binaryheirarcy.i have this problem i have to solve, im still banging my head. Ifgiven the member'Nancy' , i need to find left-most(Laura) and right-most(Robert)'Janet' = left most is ron, right most is robert'Andrew = left most is laura, right most is DavidHope u get my plan. could u help me with the sql ?
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Jul 10, 2007
Hello All,
I am going through the Data Mining Web Control sample and have that working beautifully. I now want to add in some more functionality by allowing the end user to select the: database, model and tree to view in the web viewer (similar to the parameters in SSRS). I am trying to figure out a way to get all this data programatically. So far this is what I have:
Code Snippet
For Each newDB As Microsoft.AnalysisServices.Database In myserver.Databases
dbName = newDB.Name.ToString
myLog.AppendText("DATABASE NAME - " + dbName + vbCrLf)
For Each newStructure As Microsoft.AnalysisServices.MiningStructure In newDB.MiningStructures
structureName = newStructure.Name.ToString
myLog.AppendText(" STRUCTURE NAME [ " + structureName + " ]" + vbCrLf)
For Each newModel As Microsoft.AnalysisServices.MiningModel In newStructure.MiningModels
modelName = newModel.ToString
modelAlgorithm = newModel.Algorithm.ToString
myLog.AppendText(" MODEL NAME -" + modelName + " " + modelAlgorithm + vbCrLf)
This works beautifully but I am having a problem getting the tree name which i believe its called the "attribute_name".
I can run the following SQL against Analysis Services and get the attribute name but I would like to get all this programatically and insert into a table.
Code Snippet
SELECT attribute_name
FROM [model].content
WHERE node_type=2
I plan to run this daily and update a table. This will ultimately let me pull the structure into my data mining viewer page. This way, I will only have to have one viewer page for all models, instead of a new page for each.
Thanks,
Cam
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Sep 11, 2007
Can anyone please help me? I need to create a tree structure of my company's database server. I need to include all the tables and their description in a tree structure. The output can be in any format. Is there some tool in the MSSql that will do this for me, so I don't have to write it out manualy?? Thank you very much for any ideas!!
M.
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Jun 28, 2007
Hi,
Can we represent the Decision Tree in a programatically way in an .NET application? I understand that the outcome of a Decision Tree model can be integrated into an .NET application but not sure if we can also visualize it. Does MS SQL Server support any API to render such a tree?
Thanks a lot!
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Jul 17, 2007
Hello,
I need some advice regarding organizing packages.
I am in starting phase of a data conversion project. As my ETL target, I have to produce around 70 flat files. I am creating one package for each target file. Some target files may be a result of 2-3 packages.
I have some common packages that get called from many other packages.
Before my solution tree grows into unmanageable state, I would like packages to get organized based on their target and purpose of the package.
The ideal solution to this problem would be to organize packages in a folder structure but I couldn€™t see any option to create folders in a solution.
I also noticed that we can€™t easily put package files next to each other even after renaming packages.
I was wondering is there a better or any way to organize packages in solution tree?
Kapil
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Oct 20, 2007
I have got a lot of results like the following two nodes:
All
Existing Cases: 1035298
Missing Cases: 1604
Y = 3,214,966,177,062,520,000,000.000
a >= -0.9822378254 and < -0.7867621803
Existing Cases: 45291
Missing Cases: 17
Y = 9,491,528,329,086,450,000,000.000
Every node of the tree is as odd as this. I checked the training data and found there are 5 bad points with extraordinarily high values of Y. There are over a million points, how can these five points screw up the entire analysis.
I do have good results for other predicted parameters even though they also bad points.
Any tip?
Thanks,
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May 2, 2006
I have an example case below :
Customer Id
Debt Level
Income Level
Employment Type
Credit Risk
1
High
High
Self-Employed
Bad
2
High
High
Salaried
Bad
3
High
Low
Salaried
Bad
4
Low
Low
Salaried
Good
5
Low
Low
Self-Emplyed
Bad
6
Low
High
Self-Employed
Good
7
Low
High
Salaried
Good
My question is how to make a tree from the case above I mean what method we should use to split the tree. (Mannually counting)
I hope anyone could help me by explaining i details.Because i want to make some analysis how microsoft decision tree works exactly.So Please explain me the process to build the tree completely with the method.
Thanks a lot.
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Sep 13, 2007
I have successfully created and processed a mining structure (decision tree). However, when I try to generate the tree graph, I am getting the following error:
TITLE: Microsoft Visual Studio
------------------------------
The tree graph cannot be created because of the following error:
'Object reference not set to an instance of an object.'.
For help, click: http://go.microsoft.com/fwlink?ProdName=Microsoft%u00ae+Visual+Studio%u00ae+2005&ProdVer=8.0.50727.762&EvtSrc=Microsoft.AnalysisServices.Viewers.SR&EvtID=ErrorCreateGraphFailed&LinkId=20476
------------------------------
ADDITIONAL INFORMATION:
Object reference not set to an instance of an object. (Microsoft.AnalysisServices.Viewers)
------------------------------
BUTTONS:
OK
------------------------------
...with the following technical details
===================================
The tree graph cannot be created because of the following error:
'Object reference not set to an instance of an object.'. (Microsoft Visual Studio)
------------------------------
For help, click: http://go.microsoft.com/fwlink?ProdName=Microsoft%u00ae+Visual+Studio%u00ae+2005&ProdVer=8.0.50727.762&EvtSrc=Microsoft.AnalysisServices.Viewers.SR&EvtID=ErrorCreateGraphFailed&LinkId=20476
===================================
Object reference not set to an instance of an object. (Microsoft.AnalysisServices.Viewers)
------------------------------
Program Location:
at Microsoft.AnalysisServices.Viewers.TreeViewerBase.GetTooltipText(TreeGraphNode treeNode)
at Microsoft.AnalysisServices.Viewers.TreeViewerBase.predictionTreeComboBox_SelectedIndexAction()
Has anyone encountered this same error?
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